For a complex manifold $M$, the transition functions of the tangent bundle $T(M)$ come from the Jacobian of the changeofcoordinate maps. Does there exist a related description of the transition functions of $T^{(0,1)}$ and $T^{(1,0)}$? An example would also be nice, maybe $\mathbb{CP}^1$.
1 Answer
For a real manifold $M$ the transition functions of the tangent bundle $T(M)$ come from the Jacobian of the changeofcoordinate maps.
When $M$ is complex, it has a complex tangent bundle $T_{\mathbb{C}}M$, which can be identified with the holomorphic vector bundle $T^{(1,0)} \subset TM \otimes \mathbb{C}$. The transition functions on $T_{\mathbb{C}}M$ are given by the (complex) Jacobian of the changeofcoordinate maps, so the same is true for $T^{(1,0)}$.
Since $T^{(0,1)}$ is the complex conjugate bundle, its transition functions are the complex conjugate of the same Jacobian.

$\begingroup$ To see if I understand I am going to take the example of $\mathbb{CP}^1$. The changeofcoordinate map from $\phi_1(U_1) \subset \mathbb{C}$ to $\phi_2(U_2) \subset \mathbb{C}$ is $z \mapsto z^{1}$. The complex Jacobian of this map is $z \mapsto z^{2}$. This gives the bundle $\mathcal{O}(2)$? The conjugate is $z \mapsto \overline{z}^{1}$, which is not $\mathcal{O}(2)$ as I thought it should be. $\endgroup$ Feb 1, 2010 at 23:05

1$\begingroup$ Indeed the vector bundle $T^{(0,1)}$ is a complex vector bundle, but is it not holomorphic, rather antiholomorphic. So, no chance for it to be $\mathcal{O}(2)$. $\endgroup$ Feb 1, 2010 at 23:38

$\begingroup$ ... and a similar situation holds for $\Omega^{(1,0)}(M)$ and $\Omega^{(0,1)}(M)$? $\endgroup$ Feb 2, 2010 at 0:31

$\begingroup$ Yes but by David Speyer's answer in mathoverflow.net/questions/8484/… the two bundles are isomorphic as smmoth bundles, if not as holomorphic ones. $\endgroup$ Feb 2, 2010 at 6:27